How do you differentiate #y=5x^3(sinx)^2#?

Answer 1

Paisley Johnson

#dy/dx=5x^3sin2x+15x^2sin^2x#

#"differentiate using the "color(blue)"product rule"#

#"given "y=g(x).h(x)" then "#

#dy/dx=g(x)h'(x)+h(x)g'(x)larr" product rule"#

#g(x)=5x^3rArrg'(x)=15x^2#

#h(x)=(sinx)^2rArrh'(x)=2sinxcosx=sin2x#

#rArrdy/dx=5x^3sin2x+15x^2sin^2x#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Luke Alvarez

To differentiate ( y = 5x^3(\sin x)^2 ), you can use the product rule of differentiation. The product rule states that if you have a function that is the product of two other functions, ( u(x) ) and ( v(x) ), then the derivative of the product ( u(x)v(x) ) is given by ( u'(x)v(x) + u(x)v'(x) ). Applying this rule to the given function:

[ u(x) = 5x^3 ] [ v(x) = (\sin x)^2 ]

[ u'(x) = 15x^2 ] [ v'(x) = 2\sin x \cdot \cos x ]

Using the product rule:

[ y' = u'(x)v(x) + u(x)v'(x) ] [ y' = (15x^2)(\sin x)^2 + 5x^3 \cdot 2\sin x \cdot \cos x ] [ y' = 15x^2(\sin x)^2 + 10x^3\sin x \cos x ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.